Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1ll;
  while (b > 0) {
    if (b % 2 != 0) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return res;
}
long long ncr(long long n, long long k) {
  if (k > n - k) k = n - k;
  long long pro = 1;
  for (long long i = 0; i < k; i++) {
    pro = (pro * (n - i)) % 1000000007;
    pro /= (i + 1);
  }
  return (pro % 1000000007);
}
vector<long long> prime;
void seive() {
  vector<bool> isprime(2000001, true);
  for (int i = 2; i * i <= 2000000; i++) {
    if (isprime[i]) {
      for (int j = i * i; j <= 2000000; j += i) {
        isprime[j] = false;
      }
    }
  }
  for (int i = 2; i <= 1000000; i++) {
    if (isprime[i]) prime.push_back(i);
  }
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long x = ceil((n + 1) / 2.0);
  int i;
  cout << x << endl;
  for (i = 0; i < min(n, x); i++) {
    cout << i + 1 << " " << 1 << endl;
  }
  long long j = 2;
  for (; i < n; i++) {
    cout << x << " " << j << endl;
    j++;
  }
  return 0;
}